Z Transform MCQ [Free PDF] – Objective Question Answer for Z Transform Quiz

The entire timing sequence is divided into two parts n = 0 to ∞ and n = -∞ to 0.

Since the z-transform of the signal given in the questions contains both parts, it is called a Bi-lateral z-transform.

A discrete time LTI is BIBO stable, if and only if its impulse response h(n) is absolutely summable. That is,

\(\sum_{n = -\infty}^{\infty}|h(n)|<\infty\)

We know that, H(z) = \(\sum_{n = -\infty}^{\infty}h(n)z^{-n}\)

Let z = ejω so that |z| = |ejω| = 1

Then, |H(ejω)| = |H(ω)| = | \(\sum_{n = -\infty}^{\infty}\)h(n) e^-jωn|≤\(\sum_{n = -\infty}^{\infty}\)|h(n) e^-jωn|

= \(\sum_{n = -\infty}^{\infty}\)|h(n)|<∞

Hence, we see that if the system is stable, then H(z) converges for z = ejω. That is, for a stable discrete-time LTI system, the ROC of H(z) must contain the unit circle |z| = 1.

The ROC of the z-transform of any signal cannot contain poles. Since the value of the z-transform tends to infinity, the ROC of the z-transform does not contain poles.

Given h(n) = an(n) (|a|<1)

The z-transform of h(n) is H(z) = \(\frac{z}{z-a}\), ROC is |z|>|a|

If |a|<1, then the ROC contains the unit circle. So, the system is BIBO stable.

The ROC of the causal infinite sequence is of form |z|>r1 where r1 is the largest magnitude of poles.

According to the linearity property of z-transform, if X(z) and Y(z) are the z-transforms of x(n) and y(n) respectively then, the z-transform of [x(n)↔X(z)] is x(n)+y(n) is X(z)+Y(z).

Let us divide the given x(n) into x1(n) = 3(2n)u(n) and x2(n) = 4(3n)u(n)
and x(n) = x1(n)-x2(n)

From the definition of z-transform

X1(z) = \(\frac{3}{1-2z^{-1}}\) and X2(z) = \(\frac{4}{1-3z^{-1}}\)

So, from the linearity property of z-transform

X(z) = X1(z)-X2(z)

= > X(z) = \(\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\).

By Euler’s identity, the given signal x(n) can be written as

x(n) = sin(jω0n)u(n) = \(\frac{1}{2j}[e^{jω_0n} u(n)-e^{-jω_0n} u(n)]\)

Thus X(z) = \(\frac{1}{2j}[\frac{1}{1-e^{jω_0} z^{-1}}-\frac{1}{1-e^{-jω_0} z^{-1}}]\)

On simplification, we obtain

= > \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\).

According to the definition of Z-transform

X(z) = \(\sum_{n = -\infty}^{\infty} x(n) z^{-n}\)
= >Z{x(n-k)} = \(X^1(z) = \sum_{n = -\infty}^{\infty} x(n-k) z^{-n}\)

Let n-k = l

= > X1(z) = \(\sum_{l = -\infty}^{\infty} x(l) z^{-l-k} = z^{-k}.\sum_{l = -\infty}^{\infty} x(l) z^{-l} = z^{-k}X(z)\)

We know that \(Z{u(n)} = \frac{1}{1-z^{-1}}\)

And by the time shifting property, we have Z{x(n-k)} = z-k.Z{x(n)}

= >Z{u(n-N)} = \(z^{-N}.\frac{1}{1-z^{-1}}\)

= >Z{u(n)-u(n-N)} = \(\frac{1-z^{-N}}{1-z^{-1}}\).